Download Statistical Analysis: Tukey Method, t-Test, ANOVA, and Regression and more Assignments Systems Engineering in PDF only on Docsity! 1. (Problem 2.7) Source Degrees of Freedom Sum of Squares Mean Squares Block 2 520 260 Treatment 4 498 124.5 Residual 8 40 5 total 14 (a) All entries can be determined as above. (b) The t statistics for Tukey method are calculated below: Group A(45) B(58) C(46) D(45) E(56) A(45) 7.12 0.55 0 6.02 B(58) 6.57 7.12 1.10 C(46) 0.55 5.48 D(45) 6.02 E(56) 68.4 414.1 62.6 2 1 2 1 01.0,8,5),1)(1(, ===−− qq kbk α . By comparing the t values with 4.68, Tukey method declares that the following pairs are significantly different: A&B, A&E, B&C, B&D, C&E and D&E. (c) From (b), Tukey method declares 6 pairs of treatments to be significantly different at level 0.01. The null hypothesis of the F test is that all the treatments are same to each other. Based on the conclusion of (b), we know the null hypothesis of F test is not true at level 0.01. Hence the F test will reject the null hypothesis at level 0.01. 2. (Problem 2.8) (a) In this experiment, our objective is to compare two catalysts, A and B, with respect to yield of the chemical reaction. We can clearly determine these catalysts as the treatments in our experiment. In order to get better error term estimates and increase the power of statistical inferences, different batches are used and both treatments are applied to the raw materials (our experimental units) of each batch (within which the variation is much smaller than between them). This is highly reasonable since raw material units are known to be quite different from different batches. By blocking the batch factor (and thereby eliminating a source of variation), we attain higher power of statistical inferences about the treatments. Each batch is divided into two portions, one for catalyst A and the other one for B. The treatments should be assigned to the (homogeneous units) in a random manner. Also, the timely order of the treatment applications should be randomized. This design is a randomized block design with 2 treatments: paired comparison design. (b) The correct t-test is the paired t-test since we are using a paired comparison design. 1 2 3 4 5 6 A 9 19 28 22 18 8 B 10 22 30 21 23 12 id 1 3 2 -1 5 4 With N=6 batches we get: 33.2)4...31(*6/11 1 =+++== ∑ = N i idN d 16.2)( 1 1 2/1 1 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − = ∑ = N i id ddN s and thus the paired t-statistic 646.2/ == dpaired sdNt Compare with critical value 571.2025.0,52/,1 ==− ttN α for α=0.05 level. Since the absolute value of the t-statistic is greater than the critical value (2.646 > 2.571) the paired t-test declares the two treatments as significantly different at α=0.05. (c) 95% CI for the difference between A and B. Using our results from (b), the CI is given by: 6 16.2** 025.0,52/,1 tdN s td dN ±=± − α . So the 95% CI is: ]5971.4;0629.0[ . 3. (Problem 2.15) Remark: It would have been desirable to include the interaction effect between tape speed and laser power. However, the degrees of freedom are not sufficient for this analysis, so I am only conducting a two-way ANOVA without interaction effect.
